Intensity level $200 cm$ from a source of sound is $80 dB$. If there is no loss of acoustic power in air and intensity of threshold hearing is ${10^{ - 12}}W{m^{ - 2}}$ then, what is the intensity level at a distance of $400 cm$ from source .... $dB$
Diffcult
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(b) $I \propto \frac{1}{{{r^2}}}$
$\Rightarrow \frac{{{I_2}}}{{{I_1}}} = \frac{{r_1^2}}{{r_2^2}} = \frac{{{2^2}}}{{{{(40)}^2}}} = \frac{1}{{400}}$
==> ${I_1} = 400{I_2}$
Intensity level at point $1$, ${L_1} = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right)$
and intensity at point $2,$ ${L_2} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right)$
$\therefore$ ${L_1} - {L_2} = 10\log \frac{{{I_1}}}{{{I_2}}} = 10{\log _{10}}(400)$
==> ${L_1} - {L_2} = 10 \times 2.602 = 26$
${L_2} = {L_1} - 26 = 80 - 26 = 54\;dB$
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